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Phy 202
Your 'flow experiment' report has been received. Scroll down through the document to see any comments I might have inserted, and my final comment at the end.
** Flow Experiment_labelMessages **
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Based on your knowledge of physics, answer the following, and do your best to justify your answers with physical reasoning and insight:
As water flows from the cylinder, would you expect the rate of flow to increase, decrease or remain the same as water flows from the cylinder?
Your answer (start in the next line):
My expectations would be that the rate of flow would decrease as the water flows from the cylinder. As shown above the soft drink doesnt shoot quite as far in the third picture as it did in the first because it isnt moving as fast through the whole in the third picture.
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As water flows out of the cylinder, an imaginary buoy floating on the water surface in the cylinder would descend.
Would you expect the velocity of the water surface and hence of the buoy to increase, decrease or remain the same?
Your answer (start in the next line):
I believe because the that since the water exits at a decreasing rate of flow, I also believe that the velocity would also be decreasing in the pictures from left to right
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How would the velocity of the water surface, the velocity of the exiting water, the diameter of the cylinder and the diameter of the hole be interrelated? More specifically how could you determine the velocity of the water surface from the values of the other quantities?
Your answer (start in the next line):
You could measure the levels of the liquid at three different times during the process and also make note of the time difference. If you did this you could show your results in the form of a graph, and by using the difference quotient you would find the values of the velocities. You can also find the acceleration of the liquid by using the velocity of the exiting liquid and the diameters.
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The water exiting the hole has been accelerated, since its exit velocity is clearly different than the velocity it had in the cylinder.
Explain how we know that a change in velocity implies the action of a force?
Your answer (start in the next line):
The change in velocity would be the definition of acceleration. We know that this is the action of a force because of the equation Force is equal to mass times acceleration
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What do you think is the nature of the force that accelerates the water from inside the cylinder to the outside of the outflow hole?
Your answer (start in the next line):
The nature of the force comes from the water flowing through the hole that is smaller than the cylinder causing the acceleration of flow.
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From the pictures, answer the following and justify your answers, or explain in detail how you might answer the questions if the pictures were clearer:
Does the depth seem to be changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
Your answer (start in the next line):
From the pictures shown above, it seems that the depth is changing at a regular rate
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What do you think a graph of depth vs. time would look like?
Your answer (start in the next line):
If I am correct about the depth decreasing at a regular rate then I would see the graph looking linear with a negative slope
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Does the horizontal distance (the distance to the right, ignoring the up and down distance) traveled by the stream increase or decrease as time goes on?
Your answer (start in the next line):
The horizontal distance decreases as time goes on.
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Does this distance change at an increasing, decreasing or steady rate?
Your answer (start in the next line):
I believe that the distance decreases at an increasing rate
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What do you think a graph of this horizontal distance vs. time would look like? Describe in the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph of the horizontal distance vs. time graph would be quadratic equation that is decreasing at an increasing rate. This means that the graph will start in the upper left corner and start gradually decreasing towards the lower right end of the graph.
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You can easily perform this experiment in a few minutes using the graduated cylinder that came with your kit. If you don't yet have the lab materials, see the end of this document for instructions an alternative setup using a soft-drink bottle instead of the graduated cylinder. If you will be using that alternative, read all the instructions, then at the end you will see instructions for modifying the procedure to use a soft drink bottle.
Setup of the experiment is easy. You will need to set it up near your computer, so you can use a timing program that runs on the computer. The cylinder will be set on the edge of a desk or tabletop, and you will need a container (e.g., a bucket or trash can) to catch the water that flows out of the cylinder. You might also want to use a couple of towels to prevent damage to furniture, because the cylinder will leak a little bit around the holes into which the tubes are inserted.
Your kit included pieces of 1/4-inch and 1/8-inch tubing. The 1/8-inch tubing fits inside the 1/4-inch tubing, which in turn fits inside the two holes drilled into the sides of the graduated cylinder.
Fit a short piece of 1/8-inch tubing inside a short piece of 1/4-inch tubing, and insert this combination into the lower of the two holes in the cylinder. If the only pieces of 1/4-inch tubing you have available are sealed, you can cut off a short section of the unsealed part and use it; however don't cut off more than about half of the unsealed part--be sure the sealed piece that remains has enough unsealed length left to insert and securely 'cap off' a piece of 1/4-inch tubing.
Your kit also includes two pieces of 1/8-inch tubing inside pieces of 1/4-inch tubing, with one end of the 1/8-inch tubing sealed. Place one of these pieces inside the upper hole in the side of the cylinder, to seal it.
While holding a finger against the lower tube to prevent water from flowing out, fill the cylinder to the top mark (this will be the 250 milliliter mark).
Remove your thumb from the tube at the same instant you click the mouse to trigger the TIMER program.
The cylinder is marked at small intervals of 2 milliliters, and also at larger intervals of 20 milliliters. Each time the water surface in the cylinder passes one of the 'large-interval' marks, click the TIMER.
When the water surface reaches the level of the outflow hole, water will start dripping rather than flowing continuously through the tube. The first time the water drips, click the TIMER. This will be your final clock time.
We will use 'clock time' to refer to the time since the first click, when you released your thumb from the tube and allowed the water to begin flowing.
The clock time at which you removed your thumb will therefore be t = 0.
Run the experiment, and copy and paste the contents of the TIMER program below:
Your answer (start in the next line):
1 62.206 62.206
2 64.418 2.212
3 66.941 2.523
4 69.498 2.557
5 72.182 2.684
6 75.027 2.845
7 77.873 2.846
8 81.888 4.015
9 85.749 3.861
10 90.53 4.781
11 96.669 6.139
12 108.23 11.561
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Measure the large marks on the side of the cylinder, relative to the height of the outflow tube. Put the vertical distance from the center of the outflow tube to each large mark in the box below, from smallest to largest distance. Put one distance on each line.
Your answer (start in the next line):
.6 cm
2.2 cm
3.8 cm
5.4 cm
6.9 cm
8.5 cm
9.9 cm
11.4 cm
12.9 cm
14.4 cm
15.9 cm
17.4 cm
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Now make a table of the position of the water surface vs. clock time. The water surface positions will be the positions of the large marks on the cylinder relative to the outflow position (i.e., the distances you measured in the preceding question) and the clock times will as specified above (the clock time at the first position will be 0). Enter 1 line for each event, and put clock time first, position second, with a comma between.
For example, if the first mark is 25.4 cm above the outflow position and the second is 22.1 cm above that position, and water reached the second mark 2.45 seconds after release, then the first two lines of your data table will be
0, 25.4
2.45, 22.1
If it took another 3.05 seconds to reach the third mark at 19.0 cm then the third line of your data table would be
5.50, 19.0
Note that it would NOT be 3.05, 19.0. 3.05 seconds is a time interval, not a clock time. Again, be sure that you understand that clock times represent the times that would show on a running clock.
The second column of your TIMER output gives clock times (though that clock probably doesn't read zero on your first click), the third column gives time intervals. The clock times requested here are those for a clock which starts at 0 at the instant the water begins to flow; this requires an easy and obvious modification of your TIMER's clock times.
For example if your TIMER reported clock times of 223, 225.45, 228.50 these would be converted to 0, 2.45 and 5.50 (just subtract the initial 223 from each), and these would be the times on a clock which reads 0 at the instant of the first event.
Do not make the common error of reporting the time intervals (third column of the TIMER output) as clock times. Time intervals are the intervals between clicks; these are not clock times.
Your answer (start in the next line):
0 17.4
1.88281 15.9
4.20312 14.4
6.51562 12.9
9.03906 11.4
11.78125 9.9
14.78906 8.5
18.1875 6.9
22.04687 5.4
26.71875 3.8
33.34375 2.2
49.52343 0.6
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You data could be put into the following format:
clock time (in seconds, measured from first reading) Depth of water (in centimeters, measured from the hole)
0 14
10 10
20 7
etc. etc.
Your numbers will of course differ from those on the table.
The following questions were posed above. Do your data support or contradict the answers you gave above?
Is the depth changing at a regular rate, at a faster and faster rate, or at a slower and slower rate?
Your answer (start in the next line):
The depth is changing at a slower rate. This does contradict my earlier answers.
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Sketch a graph of depth vs. clock time (remember that the convention is y vs. x; the quantity in front of the 'vs.' goes on the vertical axis, the quantity after the 'vs.' on the horizontal axis). You may if you wish print out and use the grid below.
Describe your graph in the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph is quadratic, and it is decreasing at a decreasing rate
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caution: Be sure you didn't make the common mistake of putting time intervals into the first column; you should put in clock times. If you made that error you still have time to correct it. If you aren't sure you are welcome to submit your work to this point in order to verify that you really have clock times and not time intervals
Now analyze the motion of the water surface:
For each time interval, find the average velocity of the water surface.
Explain how you obtained your average velocities, and list them:
Your answer (start in the next line):
- 9.013
-7.952
-7.803
-7.443
-7.032
-7.032
-4.980
-5.183
-4.183
-3.257
-1.729
I obtained the average velocities by using the difference quotient in the data analysis
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Assume that this average velocity occurs at the midpoint of the corresponding time interval.
What are the clock times at the midpoints of your time intervals, and how did you obtain them? (Give one midpoint for each time interval; note that it is midpoint clock time that is being requested, not just half of the time interval. The midpoint clock time is what the clock would read halfway through the interval. Again be sure you haven't confused clock times with time intervals. Do not make the common mistake of reporting half of the time interval, i.e., half the number in the third column of the TIMER's output):
Your answer (start in the next line):
1.109
3.477
6.016
8.641
11.41
14.25
17.68
21.62
25.94
31.39
40.25
I obtained these values with the midpoint values vs. the difference quotient on data analysis
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Make a table of average velocity vs. clock time. The clock time on your table should be the midpoint clock time calculated above.
Give your table below, giving one average velocity and one clock time in each line. You will have a line for each time interval, with clock time first, followed by a comma, then the average velocity.
Your answer (start in the next line):
1.109 , -9.013
3.477 , -7.952
6.016 , -7.803
8.641 , -7.443
11.41 , -7.032
14.25 , -7.032
17.68 , -4.98
21.62 , -5.183
25.94 , -4.183
31.39 , -3.257
40.25 , -1.729
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Sketch a graph of average velocity vs. clock time. Describe your graph, using the language of the Describing Graphs exercise.
Your answer (start in the next line):
The graph is fairly linear with a few values surrounding the line. It has a positive slope
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For each time interval of your average velocity vs. clock time table determine the average acceleration of the water surface. Explain how you obtained your acceleration values.
Your answer (start in the next line):
2.293
4.747
7.329
10.03
12.83
15.97
19.65
23.78
28.67
35.82
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Make a table of average acceleration vs. clock time, using the clock time at the midpoint of each time interval with the corresponding acceleration.
Give your table in the box below, giving on each line a midpoint clock time followed by a comma followed by acceleration.
Your answer (start in the next line):
2.293, -0.4481
4.747, -0.05868
7.329, -0.1371
10.03, -0.1484
12.83, 0
15.97, -0.5983
19.65, 0.05152
23.78, -0.2315
28.67, -0.1699
35.82, -0.1725
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Answer two questions below:
Do your data indicate that the acceleration of the water surface is constant, increasing or decreasing, or are your results inconclusive on this question?
Do you think the acceleration of the water surface is actually constant, increasing or decreasing?
Your answer (start in the next line):
The data shown on my velocity vs. time graph would indicate that the acceleration was constant because it is linear. I do think that the acceleration is constant, and it would be expected to be in this particular experiment.
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If position vs. clock time is quadratic, the velocity being the derivative of position the velocity should be linear, and the acceleration being the derivative of velocity should be constant.
In the ideal case this is exactly what would be observed. The behavior of the system is in fact close to ideal, and your results are consistent with what is expected.
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Go back to your graph of average velocity vs. midpoint clock time. Fit the best straight line you can to your data.
What is the slope of your straight line, and what does this slope represent? Give the slope in the first line, your interpretation of the slope in the second.
How well do you think your straight line represents the actual behavior of the system? Answer this question and explain your answer.
Is your average velocity vs. midpoint clock time graph more consistent with constant, increasing or decreasing acceleration? Answer this question and explain your answer.
Your answer (start in the next line):
The best straight line has a slope of 0.1860 which is approximately equal to 7/39. This slope indicates that there is a steady increase in velocity, which also indicates the constant acceleration. I think this line represents the behavior of the system fairly well because we would expect the constant increase and the steady, linear slope of the velocity vs. time. The v vs. t graph is more consistent with a constant acceleration indicated by the graph being linear.
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Your instructor is trying to gauge the typical time spent by students on these experiments. Please answer the following question as accurately as you can, understanding that your answer will be used only for the stated purpose and has no bearing on your grades:
Approximately how long did it take you to complete this experiment?
Your answer (start in the next line):
I spent approximately 2.5 - 3 hours on this experiment
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You may add any further comments, questions, etc. below:
Your answer (start in the next line):
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&#This lab submission looks very good. Let me know if you have any questions. &#